What is the integral of (8x+2)sqrt(4x^2+2x) ?

The integral of (8x+2)sqrt(4x^2+2x) is 1/12 (16x^2+8x)^{3/2}+C

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integral of (8x+2)sqrt(4x^2+2x)

\int (8 x + 2) 4 x^{2} + 2 x d x

\frac{1}{12} (16 x^{2} + 8 x)^{\frac{3}{2}} + C

Solution steps

\int (8 x + 2) 4 x^{2} + 2 x d x

(8 x + 2) 4 x^{2} + 2 x = 22 (4 x + 1) 2 x^{2} + x

= \int 22 (4 x + 1) 2 x^{2} + x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 22 \cdot \int (4 x + 1) 2 x^{2} + x d x

Apply u-substitution

= 22 \cdot \int \frac{u u ^{2} - 1}{8 2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 22 \frac{1}{8 2} \cdot \int u u^{2} - 1 d u

Apply u-substitution

= 22 \frac{1}{8 2} \cdot \int v^{2} d v

Apply the Power Rule

= 22 \frac{1}{8 2} \cdot \frac{v ^{3}}{3}

Substitute back

= 22 \frac{1}{8 2} \cdot \frac{( ( 4 x + 1 ) ^{2} - 1 ) ^{3}}{3}

Simplify

22 \frac{1}{8 2} \cdot \frac{( ( 4 x + 1 ) ^{2} - 1 ) ^{3}}{3} : \frac{1}{12} (16 x^{2} + 8 x)^{\frac{3}{2}}

= \frac{1}{12} (16 x^{2} + 8 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (16 x^{2} + 8 x)^{\frac{3}{2}} + C

\frac{d}{d x} (1 + 5 sin (x))

\frac{d}{d x} (e^{a x} cos (b x))

2 y^{^{'}} + 3 y = 0

d er i v a t i v e x \cdot 3 x^{2} + 5

a re a x^{2} - 2, x, - 2, 2

What is the integral of (8x+2)sqrt(4x^2+2x) ?
The integral of (8x+2)sqrt(4x^2+2x) is 1/12 (16x^2+8x)^{3/2}+C